How to evaluate the following limits? I was reading a proof on the evaluation of $\int_0^\infty e^{-x^2}\ dx$ without advanced techniques and stumbled upon two limits that I can't seem to crack:
$$\lim_{m\to\infty}\left(\sqrt{m}\cdot\prod_{n=1}^m\frac{2n}{2n+1}\right)=\frac{\sqrt{\pi}}2$$
$$\lim_{m\to\infty}\left(\sqrt{m}\cdot\prod_{n=2}^m\frac{2n-3}{2n-2}\right)=\frac1{\sqrt{\pi}}$$
The proof does not go into detail on how these limits were obtained, and since I wanted to understand it completely, I thought this would be the best place to ask. I have not been exposed to infinite products (only summations) and therefore I do not know which rules to apply (I feel as if they are quite similar?). In both cases, I see that an indeterminate form $0\cdot\infty$ presents its self, therefore I am guessing Hospital would be a nice approach? Any help  is appreciated! Also, my calculus book does not tackle infinite products, any suggestions on books that might give me a general outlook on the subject?
 A: Wallis's formula:
$$\frac{\pi}{2}=\prod_{n=1}^\infty \left[\frac{(2n)^2}{(2n+1)(2n-1)}\right].$$
Proof:
Weierstrass factorization of $\sin$ (You can find Euler's semi standard proof of this here) :
$$\sin(x)=x\prod_{n=1}^\infty\left(1-\frac{x^2}{n^2\pi^2}\right).$$
Plug in $x=\pi/2$ and play with the resulting fractions to get the desired result.
For your first product:
\begin{align*}
\prod_{n=1}^m\frac{2n}{2n+1}&=\frac{2\cdot 1}{2\cdot 1+1}\frac{2\cdot 2}{2\cdot 2+1}\frac{2\cdot 3}{2\cdot 3+1}\cdots \frac{2\cdot m}{2\cdot m+1}\\
&=2\cdot 1\frac{2\cdot 2}{2\cdot 1+1}\frac{2\cdot 3}{2\cdot 2+1}\cdots \frac{2\cdot m}{2\cdot (m-1)+1}\frac{1}{2m+1}\\
&=\frac{2}{2m+1}\prod_{n=2}^m\frac{2n}{2n-1}
\end{align*}
Thus:
\begin{align*}
\frac{\pi}{2}&=\lim_{m\rightarrow\infty}\prod_{n=1}^m \left[\frac{(2n)^2}{(2n+1)(2n-1)}\right]\\
&=\lim_{m\rightarrow\infty}\left(\prod_{n=1}^m\frac{2n}{2n+1}\right)\frac{1}{2}\left(\prod_{n=2}^m\frac{2n}{2n-1}\right)\\
&=\lim_{m\rightarrow\infty}\frac{1}{2}\frac{2m+1}{2}\left(\prod_{n=1}^m\frac{2n}{2n+1}\right)^2.
\end{align*}
Now just take the square-root of both sides and notice that $\sqrt{m}/\sqrt{\frac{2m+1}{2}}\rightarrow 1$
For the second question, try a similar trick by shifting the index $n\rightarrow n+2$.
A: Both of these can be obtained as a consequence of Stirling's approximation, by first rewriting all of the partial products in terms of factorials. This argument doesn't seem easier to me than the standard argument involving passing to a 2-dimensional integral. 
In general, a standard strategy for handling infinite products is to take their logarithms, producing infinite sums. 
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\lim_{m \to \infty}\pars{\root{m}\prod_{n = 1}^{m}{2n \over 2 n + 1}} =
\lim_{m \to \infty}{\root{m}\,m! \over \prod_{n = 1}^{m}\pars{n + 1/2}} =
\lim_{m \to \infty}{\root{m}\,m! \over \pars{3/2}^{\overline{m}}}
\\[5mm] = &\
\lim_{m \to \infty}{\root{m}\,m! \over \Gamma\pars{3/2 + m}/\Gamma\pars{3/2}} =
\Gamma\pars{3 \over 2}\lim_{m \to \infty}
{m^{1/2}\root{2\pi}m^{m + 1/2}\expo{-m} \over
\root{2\pi}\pars{m + 1/2}^{m + 1}\expo{-\pars{m + 1/2}}}
\\[5mm] = &\
{1 \over 2}\,\root{\pi}\lim_{m \to \infty}{\expo{1/2} \over
\bracks{1 + \pars{1/2}/m}^{\,m + 1}} = \bbx{\ds{\root{\pi} \over 2}}
\end{align}

The other one follows the same pattern.

